= Etendue concentration limit
{title2=$C\leq n_{\rm out}^2/(n_{\rm in}^2\sin^2\theta_{\rm in})$}
Suppose an extended source fills every position of the entrance <aperture> with a cone of half-<angle> $\theta_{\rm in}$, and all these <light rays> must reach a one-sided planar receiver. If the receiver accepts at most a cone of half-<angle> $\theta_{\rm out}$, its available <etendue> bounds the admitted <etendue>:
$$
n_{\rm in}^2A_{\rm in}\pi\sin^2\theta_{\rm in}
\leq n_{\rm out}^2A_{\rm out}\pi\sin^2\theta_{\rm out}.
$$
Consequently
$$
C\leq\frac{n_{\rm out}^2\sin^2\theta_{\rm out}}
{n_{\rm in}^2\sin^2\theta_{\rm in}}
\leq\frac{n_{\rm out}^2}{n_{\rm in}^2\sin^2\theta_{\rm in}}.
$$
This original derivation combines <conservation of etendue> with the exact cone <integral>; the second inequality allows the complete output <hemisphere>. It bounds capacity and does not prove that a chosen shape attains the bound.
\Image[/etendue-aperture-truncation-concentration.png]
{title=<Aperture> clipping discards <etendue>; full-flux concentration requires a larger output angular range}
{description=Left: a $6\,\mathrm{mm}$ stop transmits $36\%$ of the spatial disk with unchanged directional acceptance. Right: endpoint geometry for the ideal air-to-air $30^\circ$ input-cone bound, $C\leq4$. Interior sides are only a symbolic connection between ports, not a designed reflector or a traced ray path.}
For the $30^\circ$ input cone with $n_{\rm in}=n_{\rm out}=1$, $C\leq4$: a $10\,\mathrm{mm}$ entrance cannot deliver the entire filled bundle to a circular receiver smaller than $5\,\mathrm{mm}$ in <diameter>. If the output half-<angle> is restricted to $60^\circ$, $C\leq3$ and the receiver <diameter> must be at least $10/\sqrt3\approx5.7735\,\mathrm{mm}$. For a receiver medium with model <refractive index> $n_{\rm out}=1.5$, the full-<hemisphere> capacity bound instead gives $C\leq9$; that larger value also depends on transmitting the bundle into that medium.
Attempting $C=10$ in the air-to-air example forces rejection of some input. For the uniform <radiance> already specified, even filling the entire output <hemisphere> gives
$$
\eta=\frac{\Phi_{\rm out}}{\Phi_{\rm in}}
\leq\frac{G_{\rm out,capacity}}{G_{\rm in}}=\frac4{10}=0.4.
$$
The maximum mean <irradiance> gain remains $\eta C\leq4$. With nonuniform <radiance>, rejecting a given fraction of geometric <etendue> need not reject the same fraction of <radiant flux>.
These bounds use a filled extended-source bundle in passive reversible <geometrical optics>, comparing the same <wavelength> and the same radiation channels. An ideal <optical point source> does not fill that four-dimensional set. <Diffraction>, <optical coherence>, or coupling a <Gaussian beam> into an <optical fiber> require a <physical optics> calculation; the cone bound is not a universal guarantee for a single wave mode. Likewise, <optical gain> or <fluorescence> changes the propagation assumptions.
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